# Relationship between the ideal of an effective Cartier divisor and its invertible sheaf

Let $$X$$ be a scheme. We'll assume it's noetherian to avoid any pathologies. Let $$D$$ be an effective Cartier divisor on $$X$$. I am having trouble understanding how to go between the language of invertible sheaves, ideal sheaves, and effective Cartier divisors. I want to be able to go between the following two ideas:

1. An effective Cartier divisor as a pair $$(\mathcal{L}, s)$$ where $$\mathcal{L}$$ is an invertible sheaf and $$s$$ is a regular section (i.e a section $$s \in \Gamma(X, \mathcal{L})$$ whose corresponding morphism $$\mathcal{O}_{X} \longrightarrow \mathcal{L}$$ is injective).

2. An effective divisor as a sheaf of ideals $${I}_{D} \subseteq \mathcal{O}_{X}$$ which is locally generated by a single non-zerodivisor.

I know these two sheaves should be inverses of eachother. In particular, beginning with an ideal sheaf $$\mathscr{I}_{D}$$ as in (2), considering the dual $$\mathscr{Hom}_{\mathcal{O}_{X}}(\mathscr{I}_{D}, \mathcal{O}_{X})$$, we have an obvious choice for a regular section which is just given by the inclusion $$\mathscr{I}_{D} \hookrightarrow \mathcal{O}_{X}$$.

However, I am not as comfortable going from (1) to (2). Given an invertible sheaf and regular section $$(\mathcal{L}, s)$$, I want to define a sheaf of ideals $$\mathscr{Hom}_{\mathcal{O}_{X}}(\mathcal{L}, \mathcal{O}_{X})$$. The problem is I don't see any obvious way to realise this as a subsheaf of $$\mathcal{O}_{X}$$.

The obvious choice is to define a morphism $$\mathscr{Hom}_{\mathcal{O}_{X}}(\mathcal{L}, \mathcal{O}_{X}) \rightarrow \mathcal{O}_{X}$$ by evaluation at the section $$s$$ of $$\mathcal{L}$$ but I see no reason that morphism should be injective.

Is there any way to see this easily so I can translate between the two ideas?

• From (1) to (2): you may try tensoring with $\mathcal L^{-1}$ on $0\to \mathcal O_X\to \mathcal L$ Commented Aug 13, 2020 at 1:42

Suppose $$U\subseteq X$$ is an open subset on which $$\mathcal{L}$$ is trivial. It is enough to show that the map \begin{align*} e(U) : \mathscr{H}om(\mathcal{L},\mathcal{O})(U)&\to\mathcal{O}(U)\\ \phi &\mapsto \phi\circ s(1) \end{align*} is injective for any such $$U.$$ (By the way, make sure you convince yourself that this is indeed enough -- a morphism of sheaves $$\mathcal{F}\to\mathcal{G}$$ on a space $$X$$ such that $$\mathcal{F}(U)\to\mathcal{G}(U)$$ is injective for all $$U$$ in some open cover of $$X$$ need not be an injective morphism of sheaves!)
On $$U,$$ we have $$\mathscr{H}om(\mathcal{L},\mathcal{O})(U) =\operatorname{Hom}(\left.\mathcal{L}\right|_U,\mathcal{O}_U) \cong\operatorname{Hom}(\mathcal{O}_{U},\mathcal{O}_{U})\cong\mathcal{O}(U).$$ The final isomorphism here is given by $$\phi\mapsto\phi(1).$$
So, now we're looking at a map $$\mathcal{O}(U)\to\mathcal{O}(U),$$ and to describe such an $$\mathcal{O}(U)$$-module map, it suffices to specify where $$1$$ goes. Under the isomorphism $$\operatorname{Hom}(\mathcal{O}_{U},\mathcal{O}_{U})\cong\mathcal{O}(U),$$ $$1$$ corresponds to the identity morphism. So, we must compute $$e(U)(\operatorname{id}),$$ which is the image of $$1\in\mathcal{O}(U)$$ under the composition $$\mathcal{O}(U)\xrightarrow{s}\mathcal{L}(U)\cong\mathcal{O}(U)\xrightarrow{\operatorname{id}}\mathcal{O}(U).$$ Let $$f$$ be the image of $$s(1)\in\mathcal{L}(U)$$ in $$\mathcal{O}(U)$$ under the isomorphism $$\mathcal{L}(U)\cong\mathcal{O}(U).$$ Then we have $$e(U)(\operatorname{id}) = f.$$ Tracing everything out, this means that the map $$e(U) : \mathcal{O}(U)\to\mathcal{O}(U)$$ is nothing more than multiplication by $$f.$$ Now here's the key -- $$s$$ being injective means that $$s(1) = f\in\mathcal{O}(U)$$ is a nonzero divisor, so the map $$e(U)$$ is injective, as desired!
As Samiron suggests in the comment above, this may all be more simply put as follows. Recall that $$\mathscr{H}om(\mathcal{L},\mathcal{O}) = \mathcal{L}^{-1},$$ and that $$\mathcal{L}^{-1}\otimes\mathcal{L}\cong\mathcal{O}.$$ Then the map you define $$\mathscr{H}om(\mathcal{L},\mathcal{O})\to\mathcal{O}$$ is nothing more than the map you obtain by tensoring the given regular section $$s : \mathcal{O}\to\mathcal{L}$$ by $$\mathcal{L}^{-1}.$$ This produces a map $$\mathcal{L}^{-1}\to\mathcal{O},$$ which is still injective because $$\mathcal{L}^{-1}$$ is locally free (hence flat).